They actually started repeating in sequence. Guardians 125 was a while ago tho, I got 151 and there were no repeats.

That's interesting. It may indicate that there is/was a random seed associated with each account.

_________________...
Not enough time on my hands anymore...
...

WoodyGrandmaster Poster

Joined: Nov 01, 2005
Posts: 989
Location: Lake Powell

Posted:
Mon Nov 06, 2006 10:40 pm

_Indimar_ wrote:

Pulled a good one today.

14-11-12-4-4-14-11

A fine dixie.

15-11-12-10-10-15-11 PL

_________________...
Not enough time on my hands anymore...
...

MasterBeruAdvanced Master Poster

Joined: Nov 06, 2006
Posts: 368
Location: Atlanta, GA

Posted:
Tue Nov 07, 2006 1:54 pm

In a morbid way, it is funny to read your posts. I am considering returning to the arena after quite a while and I can honestly say that since the late 80s when I first began playing this has always been the complaint. And the suggestions have essentially been the same.

Hmmm... Also brings to mind why a number of us kicked around creating a similar game...

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?

Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out!

Oh well, time to dig out the discrete math text...

If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.

This (in bold) would be how many 'different sheets' if the same 5 rollups, listed in a different order, were considered a 'different sheet'(a permutation).
The number of different sheets, if the order of list is ignored and redundant rollups are disallowed, is a combination:
(woody's answer)! / 5!(woody's answer - 5)!

Concerning "woody's answer"...
The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,
-with no more than 17(oops! should be 1 elements in any container.
(something tells me there's gotta be a better way)

I haven't found any ready-made function to ram this through.

I should be able to find or make a function to do everything before the comma. Concerning the part after, it's kind of a 'negative space' problem to take away the disallowed cases. So far (in my head ), I've got about 18 cases of combo/permu functions which might(I'm probably forgetting something) do the onion-peeling/onion-wrapping. My intiution tells me that this should be reducible to 1 or 2 functions.

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?

Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out!

Oh well, time to dig out the discrete math text...

If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.

This (in bold) would be how many 'different sheets' if the same 5 rollups, listed in a different order, were considered a 'different sheet'(a permutation).
The number of different sheets, if the order of list is ignored and redundant rollups are disallowed, is a combination:
(woody's answer)! / 5!(woody's answer - 5)!

Concerning "woody's answer"...
The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,
-with no more than 17 elements in any container.
(something tells me there's gotta be a better way)

I haven't found any ready-made function to ram this through.

I should be able to find or make a function to do everything before the comma. Concerning the part after, it's kind of a 'negative space' problem to take away the dissallowed cases. So far (in my head ), I've got about 18 cases of combo/permu functions which might(I'm probably forgetting something) do the onion-peeling/onion-wrapping. My intiution tells me that this should be reducible to 1 or 2 functions.

I think you'll find yourself playing with a few sets of double integrals there, this is one of the fun project type problems.

WoodyGrandmaster Poster

Joined: Nov 01, 2005
Posts: 989
Location: Lake Powell

Posted:
Wed Nov 08, 2006 2:22 pm

pipthetroll wrote:

I think you'll find yourself playing with a few sets of double integrals there, this is one of the fun project type problems.

...in younger years, I never would have considered doing such things on a day off...but I guess this is what age and boredom does...

_________________...
Not enough time on my hands anymore...
...

DekeAdvanced Master Poster

Joined: Aug 15, 2006
Posts: 390
Location: Atlanta Georgia Area

Posted:
Wed Nov 08, 2006 3:16 pm

How Deke would generate rollups

Assuming that we would want to have a full assortment of sizes, I would first run a subroutine to determine size. Since I am not Ed Schooner and prefer d100 to 3d6 I would create the distribution I wanted and assign those values to a d100 array (since modern PC can handle data in arrays, and this is not 1983.)

A) Roll a random d100 and assign size as per the following table. We can argue about the distribution of sizes later.

I think Ed rolled 3d8 and rerolled values of 8. That would be an Ed Method.

I would then use a snowfall method for distributing the remaining points.

Distributing 51 to 69 additional points in 6 areas

B) Choose a random 1 in 6 area. If this area is less than 18 then add 4 to the area. If it is 17 or more, choose a different area. Perform this action 5 times (to distribute 20 points.)
C) Choose a random 1 in 6 area. If this area is less than 19 then add 3 to the area. If it is 18 or more, choose a different area. Perform this action 5 times (to distribute 15 points. 35 points total.)
D) Choose a random 1 in 6 area. If this area is less than 20 then add 2 to the area. If it is 19 or more, choose a different area. Perform this action 5 times (to distribute 10 points. 45 points total.
E) If the total of the 6 random area + size is less than 72, Choose a random 1 in 6 area. If this area is less than 21 then add 1 to the area. If it is 21, choose a different area. Perform this action again until total of 6 random areas + size = 72

If Wit + Will < 17 then discard and generate a new candidate. I would argue for < 18 but the group seems to have chosen 17 as the rejection point.

A true snowflake patter would add 1 to random areas many times, however that generates too flat (too boring) a distribution. Adding larger amounts to a single area generates a more varied distribution.

_________________Deke is a relic of the past known as Doc LeGrand
Arena 21, 81, 102 as Doc LeGrand's Lab

Just got a Size 19, 6 Wit, 3 Will "god" back from the Dark Arena. Does anyone want to buy this warrrior? I'll start the bidding @ $100.00...

WoodyGrandmaster Poster

Joined: Nov 01, 2005
Posts: 989
Location: Lake Powell

Posted:
Sun Nov 12, 2006 10:35 am

Woody wrote:

Maximillion wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?

The problem can be reduced descriptively to:
-How many ways are there to put 49 indistinct elements into 7 distinct containers,-with no more than 17 elements in any container.
(something tells me there's gotta be a better way)

I thought about this part some.
Looking at it from a different direction, it's just a combination w/repetition:
55! / 7! (48!)

I pretty sure how to get this part, but it will take more than scribbling on a Subway sandwich napkin to figure out.

note: these calculations involve 7 stats, not 6 as in the original post

_________________...
Not enough time on my hands anymore...
...

WoodyGrandmaster Poster

Joined: Nov 01, 2005
Posts: 989
Location: Lake Powell

Posted:
Sun Nov 12, 2006 2:23 pm

pipthetroll wrote:

Woody wrote:

Maximillian wrote:

800 + sheets okay, but does anyone know how many different warriors it is possible to design with 6 stats with #'s between 3-21, surely more than 800?

Lotsa factorials involved in that calculation.

Now I'm not gonna be able to let that one go until I figure it out!

Oh well, time to dig out the discrete math text...

If it was 7 stats from 3-21 with no 70 point limit, it would be 18^7. That would be just the possible combinations of them, not their probbility of occuring.With the limit it becomes complex enough that I'll wait for woody to figure it out. I'd guess its over a million.

Possible # of team sheets would be (woody's answer)^5, quite a few more than 856.